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postnew [5]
4 years ago
9

How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?

Chemistry
2 answers:
Andrej [43]4 years ago
7 0

Answer: -

C. The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

The kinetic energy of gas molecules increase with the increase in the temperature of the gas. With the increase in kinetic energy, the gas molecules also move faster. Thus with the increase of temperature, the speed of the molecules increase.

Temperature of first hydrogen gas sample is 10 °C.

10 °C means 273+10 = 283 K

Thus first sample temperature = 283 K

The second sample temperature of the hydrogen gas  is 350 K.

Thus the temperature is increased.

So both the kinetic energy and speed of molecules is more for the hydrogen gas sample at 350 K.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Hence the answer is C.

MrMuchimi4 years ago
7 0

Answer:

c. the hydrogen at 10 degrees celsius has lower nuclear energy than the sample at 350 k.

Explanation:

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Having energy in the reactant lead to an exothermic reaction.<br>true<br>false​
Ostrovityanka [42]

Answer:

false

Explanation:

first of all;-energy lead to an indotermic reaction.

indotermic is a reaction that absorbs energy \

*it has positive enthalpy of reaction

*Heat content of product is greater than that of reactant

*Heat is added to reactant side

example;- CO^2+2H^2+891kj --------- CH4 +2O2

6 0
3 years ago
If 10.0g of powdered iron is heated with 10.0g of sulfur in an open crucible, what is the mass of iron (II) sulfide that is form
Ugo [173]

Answer:

See Explanation

Explanation:

                     8Fe        +        S₈                =>        8FeS

Given:            10g                  10g

moles      10g/56g/mol     10g/256g/mol

                = 0.179mol Fe   = 0.039mol S₈

Reduce => divide mole values by respective coefficients; smaller value is Limiting Reactant.

                 0.179/8 = 0.022      0.039/1 = 0.039  

       => Fe is limiting reactant

                       8Fe        +               S₈                   =>         8FeS    

Given:     10g/56g/mol          10g/256g/mol

               = 0.179mol              = 0.039 mol                  0.179mol FeS produced

                                          1/8(0.179)mol S₈ used        (coefficients are equal,

                                          = 0.022 mol S₈ used       => moles Fe = moles FeS)

                                         = (0.039 - 0.022)mol S₈     = 0.179mol FeS

                                         remains in excess              =(0.179mol)(88g/mol)

                                         = 0.0166 mol S₈ (excess)  = 15.8 g FeS  

                                         = (0.0166mol)(256g/mol)           (Theoretical Yield)

                                         = 4.26g S₈ in excess

7 0
3 years ago
WILL MARK BRAINLIEST!!<br><br><br> How is a map an example of a model?
bonufazy [111]

A model represents a place thing or idea. A made is a model as it shows a place and represents the place it’s showing such as a globe

7 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
Which of the following descriptions properly describes the electrons?
gladu [14]

Answer:

Which following? If you could please be more specific I would gladely love to answer your question. (:

5 0
4 years ago
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